Answer:
The coil radius of other generator is 5.15 cm
Explanation:
Consider the equation for induced emf in a generator coil:
EMF = NBAω Sin(ωt)
where,
N = No. of turns in coil
B = magnetic field
A = Cross-sectional area of coil = π r²
ω = angular velocity
t = time
It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.
Therefore, EMF for both coils or generators will be:
EMF₁ = NBπr₁²ω Sin(ωt)
EMF₂ = NBπr₂²ω Sin(ωt)
dividing both the equations:
EMF₁/EMF₂ = (r₁/r₂)²
r₂ = r₁ √(EMF₂/EMF₁)
where,
EMF₁ = 1.8 V
EMF₂ = 3.9 V
r₁ = 3.5 cm
r₂ = ?
Therefore,
r₂ = (3.5 cm)√(3.9 V/1.8 V)
<u>r₂ = 5.15 cm</u>
Answer:
Radius of the circle will be 2.5 m
Explanation:
We have given velocity of particle moving in the circle v = 5 m/sec
Acceleration of particle in the circle 
We have to find the radius of the circle
We know that acceleration is given by 
So 
So radius of the circle will be 2.5 m
Answer:
A super conductor is a perfect conductor that has zero resistance. It doesn't just have very low resistance and conducts electricity well, it has ZERO resistance and conducts electricity perfectly with no losses at all
Answer: The hottest star is Archenar( blue) and the coolest star is Betelgeuse
Explanation:
Objects emit radiation that depends exclusively on their temperature. At an ambient temperature, the radiation emitted by an object is in the infrared spectrum (we could only see it with a special camera). If we heat it we will see that it first turns red (whose state we call “red hot”) because it is the lowest and least energetic wavelength of all.
If we continue to heat it, the wavelength that it emits to one with more energy will continue to increase and we will see that it turns yellow and then white. This is a signal that is emitting at all frequencies (but mainly in blue).
If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).
